LeetCode 1441. Build an Array With Stack Operations

Description

https://leetcode.com/problems/build-an-array-with-stack-operations/

Given an array target and an integer n. In each iteration, you will read a number from  list = {1,2,3..., n}.

Build the target array using the following operations:

  • Push: Read a new element from the beginning list, and push it in the array.
  • Pop: delete the last element of the array.
  • If the target array is already built, stop reading more elements.

Return the operations to build the target array. You are guaranteed that the answer is unique.

Example 1:

Input: target = [1,3], n = 3
Output: ["Push","Push","Pop","Push"]
Explanation: 
Read number 1 and automatically push in the array -> [1]
Read number 2 and automatically push in the array then Pop it -> [1]
Read number 3 and automatically push in the array -> [1,3]

Example 2:

Input: target = [1,2,3], n = 3
Output: ["Push","Push","Push"]

Example 3:

Input: target = [1,2], n = 4
Output: ["Push","Push"]
Explanation: You only need to read the first 2 numbers and stop.

Example 4:

Input: target = [2,3,4], n = 4
Output: ["Push","Pop","Push","Push","Push"]

Constraints:

  • 1 <= target.length <= 100
  • 1 <= target[i] <= n
  • 1 <= n <= 100
  • target is strictly increasing.

Explanation

Track the numbers added and compare with the target array to decide whether “push” or “pop” operation to use.

Python Solution

class Solution:
    def buildArray(self, target: List[int], n: int) -> List[str]:
        results = []
        numbers = []
        
        for i in range(1, n + 1):
            if numbers == target:
                return results
            
            numbers.append(i)
            
            if numbers[len(numbers) - 1] == target[len(numbers) - 1]:
                results.append("Push")
            else:
                results.append("Push")                
                results.append("Pop")
                numbers.pop()
                
        return results
                
  • Time Complexity: O(N).
  • Space Complexity: O(N).

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